Question

Find the quadratic polynomial whose zeroes are $3$ and $-4$ ?

Answer 1

Hint: Here, we are given that the two zeros or roots of a polynomial are $3$ and $-4$ . We are required to find out this polynomial. We first assume the polynomial as $a{{x}^{2}}+bx+c$ . The sum of the roots is $-\dfrac{b}{a}$ and the product is $\dfrac{c}{a}$ . Equation these values by putting the roots in these equations, and finally into the quadratic polynomial, we get the polynomial.

Complete step by step answer:
In this problem, we are given that the roots of a quadratic polynomial are $3$ and $-4$ . We need to find out this polynomial. Now, let the polynomial be $a{{x}^{2}}+bx+c$ . We know by the Sridhar Acharya formula that the roots of a polynomial can be found out with the help of the coefficients of the polynomial. The formula for this is given by,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where, “a”, “b”, and “c” have their usual meanings as that in the polynomial that we have assumed in this question.
The sum of the roots will then be, if ${{x}_{1}},{{x}_{2}}$ are the roots,
\[\begin{align}
  & {{x}_{1}}+{{x}_{2}}=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}+\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow {{x}_{1}}+{{x}_{2}}=-\dfrac{b}{a}....\left( i \right) \\
\end{align}\]
The product of the roots will then be,
 $\begin{align}
  & {{x}_{1}}{{x}_{2}}=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\times \dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow {{x}_{1}}{{x}_{2}}=\dfrac{{{b}^{2}}-\left( {{b}^{2}}-4ac \right)}{4{{a}^{2}}}=\dfrac{c}{a}....\left( ii \right) \\
\end{align}$
We know that the roots of the polynomial are $3$ and $-4$ . So, substituting them in the two equations, we get,
\[\begin{align}
  & \Rightarrow 3-4=-\dfrac{b}{a} \\
 & \Rightarrow 1=\dfrac{b}{a}....\left( iii \right) \\
\end{align}\]
 $\begin{align}
  & \Rightarrow 3\left( -4 \right)=\dfrac{c}{a} \\
 & \Rightarrow -12=\dfrac{c}{a}....\left( iv \right) \\
\end{align}$
Now, using equations (iii) and (iv), we get the polynomial as,
$a{{x}^{2}}+ax-12a=a\left( {{x}^{2}}+x-12 \right)$
Taking the value of “a” as $1$ , we get the polynomial as ${{x}^{2}}+x-12$ .

Note: We can solve the problem in another way as well. Since, we are given the two roots of the polynomial, we can find out the factors of the polynomial. The roots being $3,-4$ , the factors will be $x-3,x+4$ . The polynomial will be thus, $\left( x-3 \right)\left( x+4 \right)={{x}^{2}}+x-12$ .